In the reaction
`2CuSO_(4)+4KI rarr 2Cu_(2)I_(2)+I_(2)+2K_(2)SO_(4)` the equivalent weight of `CuSO_(4)` will be:

Options

A. 79.75
B. 159.5
C. 329
D. None of these

Explanation

To find the equivalent weight of \( CuSO_4 \) in the given reaction:

\[ 2CuSO_4 + 4KI \rightarrow 2Cu_2I_2 + I_2 + 2K_2SO_4 \]

we need to follow these steps:

### Step 1: Determine the molecular weight of \( CuSO_4 \)

The molecular weight of \( CuSO_4 \) can be calculated by adding the atomic weights of its constituent elements:

- Copper (Cu): 63.5 g/mol
- Sulfur (S): 32 g/mol
- Oxygen (O): 16 g/mol (and there are 4 oxygen atoms)

So, the calculation will be:

\[ \text{Molecular weight of } CuSO_4 = 63.5 + 32 + (4 \times 16) \]

Calculating the oxygen part:

\[ 4 \times 16 = 64 \]

Now, adding all together:

\[ \text{Molecular weight of } CuSO_4 = 63.5 + 32 + 64 = 159.5 \text{ g/mol} \]

### Step 2: Determine the n-factor of \( CuSO_4 \)

In the reaction, \( CuSO_4 \) is reduced to \( Cu_2I_2 \). The copper ion \( Cu^{2+} \) is reduced to \( Cu^0 \).

- The change in oxidation state for \( Cu^{2+} \) to \( Cu^0 \) is a change of 2 (from +2 to 0).
- Since there are 2 moles of \( CuSO_4 \) in the reaction, the total change in electrons is \( 2 \times 2 = 4 \).

Thus, the n-factor for \( CuSO_4 \) is 4 (since it provides 4 electrons).

### Step 3: Calculate the equivalent weight of \( CuSO_4 \)

The equivalent weight is calculated using the formula:

\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n \text{-factor}} \]

Substituting the values we found:

\[ \text{Equivalent weight of } CuSO_4 = \frac{159.5}{4} \]

Calculating this gives:

\[ \text{Equivalent weight of } CuSO_4 = 39.875 \text{ g/equiv} \]

### Conclusion

Thus, the equivalent weight of \( CuSO_4 \) is approximately 39.88 g/equiv.

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